Does the sum:
$$\sum_{k=1}^{\infty}k(p^{\frac{(k-1)k}{2}}-p^{\frac{(k+1)k}{2}})$$ $$ p\in\mathbb{R}|0{\leq}p<1$$ converse, and if so, to what function?
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2I don't know if this is of much help, but numerically this series seems to be the same as $\sum_{k=0}^\infty p^{\frac{(k+1)k}{2}}$, for all $0<p<1$. – 77474 May 26 '13 at 11:58
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1With @Carl's simplification this returns a theta function. – Raymond Manzoni May 26 '13 at 12:02
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1Ilya: no, I mean that your series is numerically equal to the one I posted above. Another way to write the sum I posted is, as @Raymond mentioned, as a theta function $\vartheta_2(0,\sqrt{p})/2p^{1/8}$. – 77474 May 26 '13 at 12:07
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@CarlNajafi, the terms look like they should telescope to what you state. – vonbrand May 26 '13 at 13:07
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@vonbrand: Yes, see Raymond's answer below. – 77474 May 26 '13 at 13:10
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Let's simplify your expression to get Carl Najafi's expression :
\begin{align} \sum_{k=1}^{\infty}k\left(p^{\frac{(k-1)k}{2}}-p^{\frac{(k+1)k}{2}}\right)&=\sum_{k=1}^{\infty}k\;p^{\frac{(k-1/2)^2}2-\frac 18}-\sum_{k=1}^{\infty}k\;p^{\frac{(k+1/2)^2}2-\frac 18}\\ &=\sum_{k=0}^{\infty}(k+1)\;p^{\frac{(k+1/2)^2}2-\frac 18}-\sum_{k=1}^{\infty}k\;p^{\frac{(k+1/2)^2}2-\frac 18}\\ &=\sum_{k=0}^{\infty}\;p^{\frac{(k+1/2)^2}2-\frac 18}\\ &=\sum_{k=0}^{\infty}\;p^{\frac{k(k+1)}2}\\ \end{align} Proving Carl's claim.
After that you'll simply have to use the definition of the second theta function $$\theta_2(0,\sqrt{p})=2\sum_{k=0}^{\infty}\;\sqrt{p}^{(k+1/2)^2}$$ to get the Alpha result (since $\sqrt{p}^{1/4}=\sqrt[8]{p}$) : $$\frac {\theta_2(0,\sqrt{p})}{2\;\sqrt[8]{p}}\quad\text{for}\ 0<p<1$$
Raymond Manzoni
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Thanks @Mhenni even if at least half of the credit belongs to Carl Najafi ! – Raymond Manzoni May 26 '13 at 12:33